Answer:
Step-by-step explanation:
4(x+3)+2x = 6(x+2) → 4x +12 +2x = 6x + 12 → No solution
5 + 2(3+2x) = x + 3(x+1) → 5 + 6 + 4x = x + 3x + 3 → No solution
5(x + 3) + x = 4(x+3) + 3 → 5x + 15 + x = 4x +12 +3 → 6x + 15 = 4x + 15 → x= 0
4 + 6(2+x) = 2(3x +8) → 4 + 12 + 6x = 6x + 16 → No solution
Answer:D
Step-by-step explanation:
Answer:

Step-by-step explanation:
For the sake of clarity, assuming you meant:

1) Absolute Value or Modulus functions has one property that assures us that:

2) Solving for x

Not defined in Real Set of Numbers
Evaluating
:

3) So, since for the first case the Discriminant Δ <0, then the solutions presented for
. The only solution in the Real Set for the inequality
is
, i.e. x=-1.
That is in standard form.